Find the area of the region enclosed by the curves $y = 4 x - x^{2}, y = 5 - 2 x$

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Solution

The point of intersection of the parabola and the line is given by
$4 x - x^{2} = 5 - 2 x$
$x^{2} - 6 x + 5 = 0$
$(x - 1) (x - 5) = 0$
Hence $x = 1$ and $x = 5$ .
Therefore, the area bounded by the parabola and the line between $x = 1$ and $x = 5$ is given as
$A = \int_{1}^{5} (5 - 2 x) - (4 x - x^{2}) d x$
$= \int_{1}^{5} x^{2} - 6 x + 5 d x$
$= [\frac{x^{3}}{3} - 3 x^{2} + 5 x]_{1}^{5}$
$= \frac{125}{3} - 75 + 25 - \frac{1}{3} + 3 - 5$
$= \frac{124}{3} - 50 - 2$
$= \frac{124}{3} - 52$
$= \frac{124 - 156}{3}$
$= \frac{- 32}{3}$
Hence area is $| A | = \frac{32}{3}$

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